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Topic: Finding the solubility of CaC2O4 at pH 4  (Read 16466 times)

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Offline 21385

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Finding the solubility of CaC2O4 at pH 4
« on: March 25, 2008, 12:23:31 AM »
Calculate the solubility of CaC2O4 in an aqueous solution buffered at pH 4.0.
The relevant reactions are CaC2O4=>Ca2+ + C2O4 2-     Ksp
H2C2O4=>HC2O4- + H(+)    Ka1
HC2O4- => C2O4(2-) + H(+) Ka2

I know how to obtain the solubility by using the mass balance equation, and I got the correct answer.
I balanced it like this: [Ca2+]=[C2O42-] + [HC2O4-] + [H2C2O4], solved for [Ca2+] and got the right answer

But however, when i used the charge balance equation: 2[Ca2+] + [H+] => 2[C2O4(2-)] + [HC2O4-] + [OH-], and solved for [Ca2+] but i didn't get the correct answer.

Can anyone tell me why I can't use the charge balance equation for this question?

Thanks

Offline Borek

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Re: Finding the solubility of CaC2O4 at pH 4
« Reply #1 on: March 25, 2008, 07:11:01 AM »
Show more details of your work for both cases. At this level of generality it can be everything.
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Offline JGK

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Re: Finding the solubility of CaC2O4 at pH 4
« Reply #2 on: March 25, 2008, 08:05:19 AM »
At first look, the equation


2[Ca2+] + [H+] => 2[C2O4(2-)] + [HC2O4-] + [OH-]

is not balanced for charge (I am assuming +ve and -ve charge should match).
Experience is something you don't get until just after you need it.

Offline Borek

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Re: Finding the solubility of CaC2O4 at pH 4
« Reply #3 on: March 25, 2008, 09:40:39 AM »
is not balanced for charge (I am assuming +ve and -ve charge should match).

Equation is OK. [] means concentration, it is not the same as balancing reaction equation.
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Offline JGK

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Re: Finding the solubility of CaC2O4 at pH 4
« Reply #4 on: March 25, 2008, 10:30:32 AM »
is not balanced for charge (I am assuming +ve and -ve charge should match).

Equation is OK. [] means concentration, it is not the same as balancing reaction equation.

My error, I assumed that the equation I copied which was identified as the "charge balance" equation should actually balance the positive and negative charges.

 
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Offline Borek

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Re: Finding the solubility of CaC2O4 at pH 4
« Reply #5 on: March 25, 2008, 12:02:04 PM »
I assumed that the equation I copied which was identified as the "charge balance" equation should actually balance the positive and negative charges.

Well, this is correct assumption - it should. But not in terms of molecular charges, but in terms of solution neutrality.

If you mix equimolar quantities of KCl and NaCl charge balance of the solution is

[K+] + [Na+] + [H+] = [Cl-] + [OH-]

and everything is OK, as [Cl-] = [K+] + [Na+].
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Offline 21385

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Re: Finding the solubility of CaC2O4 at pH 4
« Reply #6 on: March 25, 2008, 07:51:39 PM »
Mass Balance Method (correct answer):
[Ca2+]= [C2O4(2-)] + [HC2O4-] + [H2C2O4]
Ka1=[H+][HC2O4-]/[H2C2O4]
Ka2=[H+][C2O4(2-)]/[HC2O4-]

[Ca2+]= [C2O4(2-)] + [H+][C2O4(2-)]/Ka2 + [H+]^2[C2O4(2-)]/(Ka1Ka2)
Sub in values, and solve for [Ca2+]
[H+]=1E-4
Ksp=1.30E-8
Ka1=5.60E-2
Ka2=5.42E-5
[Ca2+]=1.92E-4 M

Charge Balance method: (wrong answer?)
2[Ca2+] + [H+] => 2[C2O4(2-)] + [HC2O4-] + [OH-]
2[Ca2+] + [H+] => 2(Ksp/[Ca2+]) + [H+][C2O4(2-)]/Ka2 + Kw/[H+]
2[Ca2+] + [H+] => 2(Ksp/[Ca2+]) + [H+]Ksp/(Ka2[Ca2+]) + Kw/[H+]
put everything in numerical values, and times every term by [Ca2+], with Kw=1.0E-14
2[Ca2+]^2 + 1E-4[Ca2+] - 5E-8 =0
Quadratic formula
[Ca2+]=1.35E-4 M
which is the wrong answer


Offline Borek

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Re: Finding the solubility of CaC2O4 at pH 4
« Reply #7 on: March 25, 2008, 08:02:28 PM »
2[Ca2+] + [H+] => 2[C2O4(2-)] + [HC2O4-] + [OH-]

I wonder... this is not a complete equation, your solution was buffered at pH 4, so it contains other ions as well. They don't have to balance by themselves - when your oxalate was dissolved, it has disturbed original composition of the solution.
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Offline 21385

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Re: Finding the solubility of CaC2O4 at pH 4
« Reply #8 on: March 25, 2008, 08:51:03 PM »
Right Right, thank you guys

I forgot about the buffer part....and since it is impossible to find out the ratio of base to acid of the buffer, we can not use the charge balance for this question, right?


Offline Borek

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